Is tan^2(x)csc^2(x)-1=tan^2(x) ?

The answer to whether tan^2(x)csc^2(x)-1=tan^2(x) is True

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prove tan^2(x)csc^2(x)-1=tan^2(x)

Solution

prove

tan^{2} (x) csc^{2} (x) - 1 = tan^{2} (x)

Solution

True

Solution steps

tan^{2} (x) csc^{2} (x) - 1 = tan^{2} (x)

Manipulating left side

tan^{2} (x) csc^{2} (x) - 1

Express with sin, cos

- 1 + csc^{2} (x) tan^{2} (x)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= - 1 + (\frac{1}{sin ( x )})^{2} tan^{2} (x)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - 1 + (\frac{1}{sin ( x )})^{2} (\frac{sin ( x )}{cos ( x )})^{2}

Simplify

- 1 + (\frac{1}{sin ( x )})^{2} (\frac{sin ( x )}{cos ( x )})^{2} : \frac{- cos ^{2} ( x ) + 1}{cos ^{2} ( x )}

- 1 + (\frac{1}{sin ( x )})^{2} (\frac{sin ( x )}{cos ( x )})^{2}

(\frac{1}{sin ( x )})^{2} (\frac{sin ( x )}{cos ( x )})^{2} = \frac{1}{cos ^{2} ( x )}

(\frac{1}{sin ( x )})^{2} (\frac{sin ( x )}{cos ( x )})^{2}

(\frac{1}{sin ( x )})^{2} = \frac{1}{sin ^{2} ( x )}

(\frac{1}{sin ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sin ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( x )}

= (\frac{sin ( x )}{cos ( x )})^{2} \frac{1}{sin ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2} = \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

= \frac{1}{sin ^{2} ( x )} \cdot \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot sin ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

Cancel the common factor:

sin^{2} (x)

= \frac{1}{cos ^{2} ( x )}

= - 1 + \frac{1}{cos ^{2} ( x )}

Convert element to fraction:

1 = \frac{1 cos ^{2} ( x )}{cos ^{2} ( x )}

= - \frac{1 \cdot cos ^{2} ( x )}{cos ^{2} ( x )} + \frac{1}{cos ^{2} ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 \cdot cos ^{2} ( x ) + 1}{cos ^{2} ( x )}

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

= \frac{- cos ^{2} ( x ) + 1}{cos ^{2} ( x )}

= \frac{- cos ^{2} ( x ) + 1}{cos ^{2} ( x )}

= \frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )}

Rewrite using trig identities

\frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )}

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - cos^{2} (x) = sin^{2} (x)

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

Rewrite using trig identities

= \frac{sin ( x )}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

\frac{sin ( x ) tan ( x )}{cos ( x )}

= tan (x) \frac{sin ( x )}{cos ( x )}

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

tan (x) tan (x)

Simplify

tan (x) tan (x) : tan^{2} (x)

tan (x) tan (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

tan (x) tan (x) = tan^{1 + 1} (x)

= tan^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= tan^{2} (x)

tan^{2} (x)

tan^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is tan^2(x)csc^2(x)-1=tan^2(x) ?
The answer to whether tan^2(x)csc^2(x)-1=tan^2(x) is True